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Chapter 5: Trigonometric Ratios Exercise – 5.2 Question: 1 sin 45° sin 30° + cos 45° cos 30° Solution: Sin 45°sin 30° + cos 45° cos 30° [1] We know that by trigonometric ratios we have, Substituting the values in equation 1, we get Question: 2 sin 60° cos 30° + cos 60° sin 30° Solution: sin 60° cos 30° + cos 60° sin 30° [1] By trigonometric ratios we have, Substituting the values in equation 1, we get Question: 3 cos 60° cos 45° – sin 60° sin 45° Solution: cos 60° cos 45° – sin 60° sin 45° [1] We know that by trigonometric ratios we have, Substituting the values in equation 1, we get Question: 4 sin230° + sin245° + sin260° + sin290° Solution: sin230° + sin245° + sin260° + sin290° [1] We know that by trigonometric ratios we have, Substituting the values in equation 1, we get Question: 5 cos230° + cos245° + cos260° + cos290° Solution: cos230° + cos245° + cos260° + cos290° [1] We know that by trigonometric ratios we have, Substituting the values in equation 1, we get Question: 6 tan230° + tan245° + tan260° Solution: tan230° + tan245° + tan260° [1] We know that by trigonometric ratios we have, Substituting the values in equation 1, we get Question: 7 2 sin230° − 3 cos245° + tan260° Solution: 2sin230° − 3cos245° + tan260° [1] We know that by trigonometric ratios we have, Substituting the values in equation 1, we get Question: 8 sin2 30° cos245°+ 4 tan230°+ 1/2sin2 90° - 2 cos2 90°+ 1/24 cos20° Solution: sin2 30° cos245°+ 4 tan230°+ 1/2sin2 90° - 2 cos2 90°+ 1/24 cos20° [1] We know that by trigonometric ratios we have, Substituting the values in equation 1, we get Question: 9 4(sin460° + cos430°) − 3(tan260° − tan245°) + 5cos245° Solution: 4(sin460° + cos430°) − 3(tan260° − tan245°) + 5cos245° [1] We know that by trigonometric ratios we have, Substituting the values in equation 1, we get Question: 10 (cosec2 45° sec2 30°) (sin230° + 4 cot245° − sec2 60°) Solution: (cosec245° sec230°)(sin230° + 4 cot245° − sec260°) [1] We know that by trigonometric ratios we have, Substituting the values in equation 1, we get Question: 11 cosec3 30°cos 60° tan3 45° sin2 90° sec2 45°cot 30° Solution: = cosec3 30°cos 60° tan3 45° sin2 90° sec2 45° cot 30° Question: 12 cot230° − 2cos260° − 3/4 sec245° – 4 sec230° Solution: = cot230° − 2cos260° − 3/4 sec245° – 4 sec230° Question: 13 (cos 0° + sin 45° + sin 30°) (sin 90∘ − cos 45° + cos 60°) Solution: Given, (cos 0° + sin 45° + sin 30°) (sin 90∘ − cos 45° + cos 60°) Question: 14 Solution: Given, Question: 15 Solution: Given, Question: 16 4(sin430° + cos260°) − 3(cos245° − sin290°) − sin260° Solution: Given, 4(sin430° + cos260°) − 3(cos245° − sin290°) − sin260° Question: 17 Solution: Given, = 3 + 2 + 4 = 9 Question: 18 Solution: Given, Question: 19 Solution: Question: 20 2 sin 3x = √3 Solution: Given, 2 sin 3x = √3 ⟹ sin 3x = √3/2 ⟹ sin 3x = sin 60° ⟹ 3x = 60° ⟹ x = 20° Question: 21 Solution: x/2 = 30° x = 60° Question: 22 √3 sin x = cos x Solution: √3 tan x = 1 tan x = 1/√3 ∴ tan x = tan45° x = 45° Question: 23 tan x = sin 45° cos 45° + sin 30° Solution: tan x = 1 tan x = 45° x = 45° Question: 24 √3 tan 2x = cos 60°+ sin 45°cos 45° Solution: 2x = 30° x = 15° Question: 25 cos 2x = cos 60° cos 30° + sin 60° sin 30° Solution: 2x = 30° x = 15° Question: 26 If θ = 30°, verify (iv) cos 3θ = 4cos3θ − 3cos θ cos 3θ = 4 cos3 θ – 3 cos θ Solution: Substitute θ = 30° θ = 30° in equation (i) Therefore, LHS = RHS Substitute θ = 30° θ = 30° Therefore, LHS = RHS. Substitute θ = 30°θ = 30° Therefore, LHS = RHS (iv) cos 3θ = 4cos3θ − 3cos θ cos 3θ = 4 cos3θ – 3 cos θ Solution: LHS = cos 3θ Substitute θ = 30° = cos 3(30°) = cos 90° = 0 RHS = 4 cos3θ – 3 cos θ = 4 cos330° − 3 cos 30° = 0 Therefore, LHS = RHS. Question: 27 If A = B = 60°. Verify (i) cos (A – B) = cos A cos B + sin A sin B Solution: cos (A – B) = cos A cos B + sin A sin B … (i) Substitute A and B in (i) ⟹ cos (60° – 60°) = cos 60° cos 60° + sin 60° sin 60° ⟹ 1 = 1 Therefore, LHS = RHS (ii) Substitute A and B in (i) ⟹ sin (60° – 60°) = sin 60° cos 60° – cos 60° sin 60° ⟹ sin0° = 0 ⟹ 0 = 0 Therefore, LHS = RHS A = 60°, B = 60° we get, tan 0° = 0 0 = 0 Therefore, LHS = RHS Question: 28 If A = 30°, B = 60° verify: (i) Sin (A + B) = Sin A cos B + Cos A Sin B (ii) Cos (A + B) = Cos A Cos B – Sin A Sin B Solution: (i) A = 30°, B = 60° we get Sin (30° + 60°) = Sin 30° Cos 60° + Cos 30° Sin 60° Sin (90°) = 1 ⟹ 1 = 1 Therefore, LHS = RHS (ii) Cos (A + B) = Cos A Cos B – Sin A Sin B A = 30°, B = 60° we get Cos (30° + 60°) = Cos 30° Cos 60° – Sin 30° Sin 60° 0 = 0 Therefore, LHS = RHS Question: 29 If sin(A + B) = 1 and cos(A - B) = 1, 0° < A + B ≤ 90°, A ≥ B find A and B. Solution: Given, sin(A + B) = 1 this can be written as sin (A + B) = sin (90°) sin (90°) cos(A - B) = 1 this can be written as cos (A - B) = cos (0°) cos (0°) ⟹ A + B = 90° A – B = 0° 2A = 90° A = 90°/2 A = 45° Substitute A value in A – B = 0° 45°– B = 0° B = 45° Hence, the value of A = 45°and B = 45° Question: 30 If tan (A - B) = 1/√3 and tan (A + B) = √3, 0°<A + B ≤ 90°, A > B find A and B Solution: Given, A – B = 30° ... 1 A + B = 60° ... 2 Solve equations 1 and 2 A + B = 30° A – B = 60° 2A = 90° A = 90°/2 A = 45° Substitute the value of A in equation 1 45°+ B = 30° B = 30° – 45° B = 15° The value of A = 45°and B = 15° Question: 31 If sin (A - B) = 1/2 and cos (A + B) = 1/2, 0°<A + B ≤ 90°, A < B. find A and B. Solution: Given, A + B = 60° ... 2 Solve equations 1 and 2 A + B = 60° A – B = 30° 2A = 90° A = 90°/2 A = 45° Substitute the value of A in equation 2 45°+ B = 60° B = 60° – 45° B = 15° The value of A = 45°and B = 15° Question: 32 In a Δ ABC right angled triangle at B, ∠A = ∠C. Find the values of: 1. sin A cos C + cos A sin C 2. sin A sin B + cos A cos B Solution: 1. since, it is given as ∠A = ∠C the value of A and C is 45°, the value of angle B is 90° because the sum of angles of triangle is 180° ⟹ sin 45° cos 45° + cos 45° sin 45° ⟹ 1 The value of sin A cos C + cos A sin C is 1 2. since, it is given as ∠A = ∠C the value of A and C is 45°, the value of angle B is 90° because the sum of angles of triangle is 180° ⟹ sin 45°sin 90° + cos 45°sin 90° The value of sin A sin B + cos A cos B is 1/√2 Question: 33 Find the acute angle A and B, if sin (A + 2B) = √3/2 and cos(A + 4B) = 0, A > B. Solution: Given, A + 2B = 60° ... 1 Cos (A + 4B) = 0 A + 4B = sin−1(90) sin−1(90) A + 4B = 90° ... 2 Solve equations 1 and 2 2B = 30° B = 30°/2 B = 15° Substitute B value in equation 2 A + 4B = 90° A + 4(15°) = 90° A + 60° = 90° A = 90° – 60° A = 30° The value of A = 30° and B = 15° Question: 34 In ΔPQR, right angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠ P and ∠ R. Solution: In ΔPQR, right angled at Q, PQ = 3 cm and PR = 6 cm By Pythagoras theorem, PR2 = PQ2 + QR2 ⟹ 62 = 32 + QR2 ⟹ QR2 = 36 – 9 ⟹ QR = √27 ⟹ QR = 3√3 sin R = 3/6 = 1/2 = sin30° ∠R = 30° As we know, Sum of angles in a triangle = 180 ∠P + ∠Q + ∠R = 180° ⟹ ∠P + 90° + 30° = 180° ⟹ ∠P = 180° – 120° ⟹ ∠P = 60° Therefore, ∠R = 30° And, ∠P = 60° Question: 35 If sin (A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of sin 15 and cos 15. Solution: Given, sin (A – B) = sin A cos B – cos A sin B And, cos (A – B) = cos A cos B + sin A sin B We need to find, sin 15 and cos 15. Let A = 45 and B = 30 sin 15 = sin (45- 30) = sin 45 cos 30 – cos 45 sin 30 cos 15 = cos (45- 30) = cos 45 cos 30 – sin 45 sin 30 Question: 36 In a right triangle ABC, right angled at C, if ∠B = 60° and AB = 15 units. Find the remaining angles and sides. Solution: Sin 60°= x/15 cos 60°= x/15 x = 7.5 units Question: 37 In ΔABC is a right triangle such that ∠C = 90°, ∠A = 45°and BC = 7 units. Find the remaining angles and sides. Solution: Here, ∠C = 90° and ∠A = 45° We know that, ∠A + ∠B + ∠C = 180° ⟹ 45°+ 90° + ∠C = 180° ⟹ 135° + ∠C = 180° ⟹ ∠C = 180° – 135° ⟹ ∠C = 45° The value of the remaining angle C is 45° Now, we need to find the sides x and y here, y =7√2 units x = 7 units the value of x = 7 units and y = √2 units Question: 38 In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD. Solution: Let AC = x cm and CB = y cm ⇒ x = 40 cm = AC Similarly BD = 40 cm Now, Question: 39 If A & B are acute angles such that tan A = 1/2 tan B = 1/3 and tan (A + B) =find A+B. Solution: (A + B) = Tan-1 (1) (A + B) = 45° Question: 40 Prove that: (√3 -1) (3 – cot 30°) = tan360° - 2 sin 60° Solution: R.H.S ⟹ tan3 60° - 2 sin 60° L.H.S = R.H.S Hence prove
sin 45° sin 30° + cos 45° cos 30°
Sin 45°sin 30° + cos 45° cos 30° [1]
We know that by trigonometric ratios we have,
Substituting the values in equation 1, we get
sin 60° cos 30° + cos 60° sin 30°
sin 60° cos 30° + cos 60° sin 30° [1]
By trigonometric ratios we have,
cos 60° cos 45° – sin 60° sin 45°
cos 60° cos 45° – sin 60° sin 45° [1]
sin230° + sin245° + sin260° + sin290°
sin230° + sin245° + sin260° + sin290° [1]
cos230° + cos245° + cos260° + cos290°
cos230° + cos245° + cos260° + cos290° [1]
tan230° + tan245° + tan260°
tan230° + tan245° + tan260° [1]
2 sin230° − 3 cos245° + tan260°
2sin230° − 3cos245° + tan260° [1]
sin2 30° cos245°+ 4 tan230°+ 1/2sin2 90° - 2 cos2 90°+ 1/24 cos20°
sin2 30° cos245°+ 4 tan230°+ 1/2sin2 90° - 2 cos2 90°+ 1/24 cos20° [1]
4(sin460° + cos430°) − 3(tan260° − tan245°) + 5cos245°
4(sin460° + cos430°) − 3(tan260° − tan245°) + 5cos245° [1]
Question: 10
(cosec2 45° sec2 30°) (sin230° + 4 cot245° − sec2 60°)
(cosec245° sec230°)(sin230° + 4 cot245° − sec260°) [1]
cosec3 30°cos 60° tan3 45° sin2 90° sec2 45°cot 30°
= cosec3 30°cos 60° tan3 45° sin2 90° sec2 45° cot 30°
cot230° − 2cos260° − 3/4 sec245° – 4 sec230°
= cot230° − 2cos260° − 3/4 sec245° – 4 sec230°
(cos 0° + sin 45° + sin 30°) (sin 90∘ − cos 45° + cos 60°)
Given, (cos 0° + sin 45° + sin 30°) (sin 90∘ − cos 45° + cos 60°)
Given,
4(sin430° + cos260°) − 3(cos245° − sin290°) − sin260°
= 3 + 2 + 4
= 9
2 sin 3x = √3
⟹ sin 3x = √3/2
⟹ sin 3x = sin 60°
⟹ 3x = 60°
⟹ x = 20°
x/2 = 30°
x = 60°
√3 sin x = cos x
√3 tan x = 1
tan x = 1/√3
∴ tan x = tan45°
x = 45°
tan x = sin 45° cos 45° + sin 30°
tan x = 1
tan x = 45°
√3 tan 2x = cos 60°+ sin 45°cos 45°
2x = 30°
x = 15°
cos 2x = cos 60° cos 30° + sin 60° sin 30°
If θ = 30°, verify
(iv) cos 3θ = 4cos3θ − 3cos θ cos 3θ = 4 cos3 θ – 3 cos θ
Substitute θ = 30° θ = 30° in equation (i)
Therefore, LHS = RHS
Substitute θ = 30° θ = 30°
Therefore, LHS = RHS.
Substitute θ = 30°θ = 30°
(iv) cos 3θ = 4cos3θ − 3cos θ cos 3θ = 4 cos3θ – 3 cos θ
Solution:
LHS = cos 3θ
Substitute θ = 30°
= cos 3(30°) = cos 90°
= 0
RHS = 4 cos3θ – 3 cos θ
= 4 cos330° − 3 cos 30°
If A = B = 60°. Verify
(i) cos (A – B) = cos A cos B + sin A sin B
cos (A – B) = cos A cos B + sin A sin B … (i)
Substitute A and B in (i)
⟹ cos (60° – 60°) = cos 60° cos 60° + sin 60° sin 60°
⟹ 1 = 1
(ii) Substitute A and B in (i)
⟹ sin (60° – 60°) = sin 60° cos 60° – cos 60° sin 60°
⟹ sin0° = 0
⟹ 0 = 0
A = 60°, B = 60° we get,
tan 0° = 0
0 = 0
If A = 30°, B = 60° verify:
(i) Sin (A + B) = Sin A cos B + Cos A Sin B
(ii) Cos (A + B) = Cos A Cos B – Sin A Sin B
(i) A = 30°, B = 60° we get
Sin (30° + 60°) = Sin 30° Cos 60° + Cos 30° Sin 60°
Sin (90°) = 1 ⟹ 1 = 1
A = 30°, B = 60° we get
Cos (30° + 60°) = Cos 30° Cos 60° – Sin 30° Sin 60°
If sin(A + B) = 1 and cos(A - B) = 1, 0° < A + B ≤ 90°, A ≥ B find A and B.
sin(A + B) = 1 this can be written as sin (A + B) = sin (90°) sin (90°)
cos(A - B) = 1 this can be written as cos (A - B) = cos (0°) cos (0°)
⟹ A + B = 90°
A – B = 0°
2A = 90°
A = 90°/2
A = 45°
Substitute A value in A – B = 0°
45°– B = 0°
B = 45°
Hence, the value of A = 45°and B = 45°
If tan (A - B) = 1/√3 and tan (A + B) = √3, 0°<A + B ≤ 90°, A > B find A and B
A – B = 30° ... 1
A + B = 60° ... 2
Solve equations 1 and 2
A + B = 30°
A – B = 60°
Substitute the value of A in equation 1
45°+ B = 30°
B = 30° – 45°
B = 15°
The value of A = 45°and B = 15°
If sin (A - B) = 1/2 and cos (A + B) = 1/2, 0°<A + B ≤ 90°, A < B. find A and B.
A + B = 60°
A – B = 30°
Substitute the value of A in equation 2
45°+ B = 60°
B = 60° – 45°
In a Δ ABC right angled triangle at B, ∠A = ∠C. Find the values of:
1. sin A cos C + cos A sin C
2. sin A sin B + cos A cos B
1. since, it is given as ∠A = ∠C
the value of A and C is 45°, the value of angle B is 90°
because the sum of angles of triangle is 180°
⟹ sin 45° cos 45° + cos 45° sin 45°
⟹ 1
The value of sin A cos C + cos A sin C is 1
2. since, it is given as ∠A = ∠C
⟹ sin 45°sin 90° + cos 45°sin 90°
The value of sin A sin B + cos A cos B is 1/√2
Find the acute angle A and B, if sin (A + 2B) = √3/2 and cos(A + 4B) = 0, A > B.
A + 2B = 60° ... 1
Cos (A + 4B) = 0
A + 4B = sin−1(90) sin−1(90)
A + 4B = 90° ... 2
2B = 30°
B = 30°/2
Substitute B value in equation 2
A + 4B = 90°
A + 4(15°) = 90°
A + 60° = 90°
A = 90° – 60°
A = 30°
The value of A = 30° and B = 15°
In ΔPQR, right angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠ P and ∠ R.
In ΔPQR, right angled at Q, PQ = 3 cm and PR = 6 cm
By Pythagoras theorem,
PR2 = PQ2 + QR2
⟹ 62 = 32 + QR2
⟹ QR2 = 36 – 9
⟹ QR = √27
⟹ QR = 3√3
sin R = 3/6 = 1/2 = sin30°
∠R = 30°
As we know, Sum of angles in a triangle = 180
∠P + ∠Q + ∠R = 180°
⟹ ∠P + 90° + 30° = 180°
⟹ ∠P = 180° – 120°
⟹ ∠P = 60°
Therefore, ∠R = 30°
And, ∠P = 60°
If sin (A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of sin 15 and cos 15.
sin (A – B) = sin A cos B – cos A sin B
And, cos (A – B) = cos A cos B + sin A sin B
We need to find, sin 15 and cos 15.
Let A = 45 and B = 30
sin 15 = sin (45- 30) = sin 45 cos 30 – cos 45 sin 30
cos 15 = cos (45- 30) = cos 45 cos 30 – sin 45 sin 30
In a right triangle ABC, right angled at C, if ∠B = 60° and AB = 15 units. Find the remaining angles and sides.
Sin 60°= x/15
cos 60°= x/15
x = 7.5 units
In ΔABC is a right triangle such that ∠C = 90°, ∠A = 45°and BC = 7 units. Find the remaining angles and sides.
Here, ∠C = 90° and ∠A = 45°
We know that,
∠A + ∠B + ∠C = 180°
⟹ 45°+ 90° + ∠C = 180°
⟹ 135° + ∠C = 180°
⟹ ∠C = 180° – 135°
⟹ ∠C = 45°
The value of the remaining angle C is 45°
Now, we need to find the sides x and y here,
y =7√2 units
x = 7 units
the value of x = 7 units and y = √2 units
In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.
Let AC = x cm and CB = y cm
⇒ x = 40 cm = AC
Similarly BD = 40 cm
Now,
If A & B are acute angles such that tan A = 1/2 tan B = 1/3 and tan (A + B) =find A+B.
(A + B) = Tan-1 (1)
(A + B) = 45°
Prove that: (√3 -1) (3 – cot 30°) = tan360° - 2 sin 60°
R.H.S ⟹ tan3 60° - 2 sin 60°
L.H.S = R.H.S
Hence prove
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Chapter 5: Trigonometric Ratios Exercise –...